Re: Entropy, Objectivity, and Timescales

[DSCHROEDER@cc.weber.edu]

In response to Leigh's response to my response to James...

> > > Take the initial quantum state, and propagate it with the Schrodinger
> > > eq. until there is a non-zero probability of finding the system in each
> > > state considered to be accessible.  Does that work for everyone?
> >
> > No, doesn't work for me.  Let's assume that our system is isolated,
> > and use a basis of stationary (definite-energy) states.  Then the
> > Schrodinger equation predicts that only those states that appear
> > in the initial wavefunction (decomposed in this basis) will ever appear
> > in the wavefunction in the future.  Thus, you can wait as long as you
> > like, but the number of "accessible" states (counted in this way) will
> > never increase.
>
> For heaven's sake! Eigenstates do not persist for long times in dense
> systems; they never even occur. Why do you assume an isolated system?
> We are talking about entropy, an attribute of a real system. Unphysical
> assumptions do not help clarify what is a conceptual problem.

I don't see the harm in assuming that the system (of 10^23 molecules
or whatever) is isolated from the rest of the universe.  We need to make 
some simplifying assumptions if we're to construct any models at all.  

On the other hand, I never assumed the system was in an eigenstate.
My statement is exact, for any initial state of the system.  If the
initial state is a mixture of all 10^(10^23) stationary states, then
so is the final state.  But if the initial state misses some (or most)
of them, then so does the final state.  

> What James means is that we express our microstate as a combination of
> single system states (e.g. particles in a box of gas) and allow them to
> interact. The propagation of the interaction then proceeds at some
> finite speed near sound speed.

So instead of a basis of exact stationary states for the whole system, 
you'd prefer to use a basis of products of single-particle states.
To me it's not at all obvious what happens in this basis, that is,
whether all 10^(10^23) possible states will eventually get mixed into
the wavefunction (according to the Schrodinger equation).  That's why
I chose to work in a basis where the answer is obvious.

Do you know a way to prove that all 10^(10^23) states *will* get
mixed into the final wavefunction, using your basis?  This isn't 
obvious to me at all.  And even if it's true, can you tell me why 
your basis is better than mine?  (If entropy depends on our choice
of basis, doesn't that make it subjective?)

-dan